CSCI 5551 - Introduction to Intelligent Robotics
Assignment 1 Solutions
Problem 1
(a) In this problem we are given two vectors, p′a = (0, 1, 2)
T and p′b = (−1, 1, 1)T , with
respect to a coordinate frame that has been rotated 30 degrees about the OX axis.
We are asked to find the pa and pb with respect to the original coordinate system.
pa = Rx,30 · p′a =⇒
1 0 00 cos(30) −sin(30)
0 cos(30) sin(30)

01
2
 =
 0−0.1340
2.2321

similarly for the second vector:
pb = Rx,30 · p′b =⇒
1 0 00 cos(30) −sin(30)
0 sin(30) cos(30)

−11
1
 =
 −10.336
1.366

(b) For the second part, we are given the same two vectors, but with respect to the original
coordinate system, pa = (0, 1, 2)
T and pb = (−1, 1, 1)T ; and we are asked to find their
new coordinates with respect to a frame that has been rotated around OX by 30
degrees, p′a and p
′
b. From the theory we know these equations:
pa = Rx,30·p′a =⇒ p′a = RTx,30· pa ∴ p
′
a =
1 0 00 cos(30) −sin(30)
0 sin(30) cos(30)

T 01
2
 =
 01.8660
1.2321

similarly for the second vector:
pb = Ry,60p
′
b =
1 0 00 cos(30) −sin(30)
0 sin(30) cos(30)

T −11
1
 =
 −11.366
0.366

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CSCI 5551 - Introduction to Intelligent Robotics
Problem 2
This problem considers 6 DOF planar manipulator RRRRRR, as can be seen in figure 2.
Figure 1: RRRRRR Manipulator.
The DH-Table for that manipulator is:
i αi−1 ai−1 d θ
1 0 0 0 θ1
2 0 D 0 θ2
3 0 D 0 θ3
4 0 D 0 θ4
5 0 D 0 θ5
6 0 D 0 θ6
The individual transformations from the above table are:
T01 =
(
Rz,θ1 03x1
01x3 1
)
=

cθ1 −sθ1 0 0
sθ1 cθ1 0 0
0 0 1 0
0 0 0 1

T12 =
(
I3x3 dx
01x3 1
)(
Rz,θ2 03x1
01x3 1
)
=

cθ2 −sθ2 0 D
sθ2 cθ2 0 0
0 0 1 0
0 0 0 1

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CSCI 5551 - Introduction to Intelligent Robotics
T23 =
(
I3x3 dx
01x3 1
)(
Rz,θ3 03x1
01x3 1
)
=

cθ3 −sθ3 0 D
sθ3 cθ3 0 0
0 0 1 0
0 0 0 1

T34 =
(
I3x3 dx
01x3 1
)(
Rz,θ4 03x1
01x3 1
)
=

cθ4 −sθ4 0 D
sθ4 cθ4 0 0
0 0 1 0
0 0 0 1

T45 =
(
I3x3 dx
01x3 1
)(
Rz,θ5 03x1
01x3 1
)
=

cθ5 −sθ5 0 D
sθ5 cθ5 0 0
0 0 1 0
0 0 0 1

T56 =
(
I3x3 dx
01x3 1
)(
Rz,θ5 03x1
01x3 1
)
=

cθ6 −sθ6 0 D
sθ6 cθ6 0 0
0 0 1 0
0 0 0 1

And the transformation from the base to the end-effector is:
T06 = T01T12T23T34T45T56 =
(
Rz,θ12345 D(Rz,θ1 +Rz,θ12 +Rz,θ123 +Rz,θ1234 +Rz,θ12345 +Rz,θ123456)
01x3 1
)
=

cθ123456 −sθ123456 0 D(cθ1 + cθ12 + cθ123 + cθ1234 + cθ12345 + cθ123456)
sθ123456 cθ123456 0 D(sθ1 + sθ12 + sθ123 + sθ1234 + sθ12345 + sθ123456)
0 0 1 0
0 0 0 1

Where c, s stand for cos and sin respectively, and θ12 = θ1 + θ2.
In the generalized case for the H-1 DOF of the planar manipulator, we get the following
figure and DH-Table.
i a α θ d
1 0 0 θ1 0
2 D 0 θ2 0
3 D 0 θ3 0
... ... ... ... ...
M-1 D 0 θM−1 0
M D 0 0 0
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CSCI 5551 - Introduction to Intelligent Robotics
The individual transformations from the above table are following the similar pattern
with the previous:
TM−2,M−1 =
(
I3x3 dx
01x3 1
)(
Rz,θM−1 03x1
01x3 1
)
=

cθM−1 −sθM−1 0 D
sθM−1 cθM−1 0 0
0 0 1 0
0 0 0 1

And the transformation from the base to the end-effector is:
T0,M = T01T12...TM−2,M−1TM−1,M =
(
Rz,θ12...M−1 (Rz,θ1 +Rz,θ12 +Rz,θ12...M−1)D
01x3 1
)
=

cθ12...M−1 −sθ12...M−1 0 D(cθ1 + cθ12 + cθ12...M−1)
sθ12...M−1 cθ12...M−1 0 D(sθ1 + sθ12 + sθ12...M−1)
0 0 1 0
0 0 0 1

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CSCI 5551 - Introduction to Intelligent Robotics
Problem 3
For the frame assignments of the RRP manipulator we have:
Figure 2: RRP Manipulator.
The DH-Table for that manipulator is:
i α a d θ
1 0 0 0 θ1
2 0 L 0 θ2
3 0 L 0 ω3-90
4 -90 0 L3 0
5 90 0 0 90
Since the manipulator ends with a prismatic joint, we add an extra transformation (row
5 above) in order to align the final frame with the origin frame, that is with Z pointing out
of the page.
The individual transformations from the above table are:
T01 =
(
Rz,θ1 03x1
01x3 1
)
=

cθ1 −sθ1 0 0
sθ1 cθ1 0 0
0 0 1 0
0 0 0 1

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CSCI 5551 - Introduction to Intelligent Robotics
T12 =
(
I3x3 dx
01x3 1
)(
Rz,θ2 03x1
01x3 1
)
=
(
Rz,θ2 dx
01x3 1
)
=

cθ2 −sθ2 0 L
sθ2 cθ2 0 0
0 0 1 0
0 0 0 1

T23 =
(
I3x3 dx
01x3 1
)(
Rz,ω3−90 03x1
01x3 1
)
=
(
Rz,ω3−90 dx
01x3 1
)
=

c(ω3 − 90) −s(ω3 − 90) 0 L
s(ω3 − 90) c(ω3 − 90) 0 0
0 0 1 0
0 0 0 1

T34 =
(
Rx,−90 0
01x3 1
)(
I3x3 dz
01x3 1
)
=
(
Rz,−90 Rx,−90dz
01x3 1
)
=

1 0 0 0
0 0 1 L3
0 −1 0 0
0 0 0 1

T45 =

1 0 0 0
0 0 −1 0
0 1 0 0
0 0 0 1


0 −1 0 0
1 0 0 0
0 0 1 0
0 0 0 1

So the final transformation is:
T = T01 ∗ T12 ∗ T23 ∗ T34 ∗ T45 =

c(θ12 + ω3) −s(θ12 + ω3) 0 L3c(θ12 + ω3) + L(cθ1 + cθ12)
s(θ12 + ω3) c(θ12 + ω3) 0 L3s(θ12 + ω3) + L(sθ1 + sθ12)
0 0 1 0
0 0 0 1

So, the equation for the x, y, φ of the end-effector should be:
x = L3c(θ12 + ω3) + L(cθ1 + cθ12)
y = L3s(θ12 + ω3) + L(sθ1 + sθ12)
φ = θ12 + ω3
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CSCI 5551 - Introduction to Intelligent Robotics
Problem 4
In this problem we have a series of transformations that occur at a particular sequence. We
have a rotation at the X-axis by an angle a.
T1 =

1 0 0 0
0 cosa −sina 0
0 sina cosa 0
0 0 0 1

Prior to that, we have a translation along to Z-axis by b units.
T2 =

1 0 0 0
0 1 0 0
0 0 1 b
0 0 0 1

Finally, we have a rotation of φ angle on the new Y-axis.
T3 =

cosφ 0 sinφ 0
0 1 0 0
−sinφ 0 cosφ 0
0 0 0 1

These are all the given transformation. Now, in order to find the final T we need to
multiply them in the correct order.
T = T2 ∗ T1 ∗ T3 =
cos(φ) 0 sin(φ) 0
sin(a)sin(φ) cos(a) −sin(a)cos(φ) 0
−sin(φ)cos(a) sin(a) cos(φ)cos(a) d
0 0 0 1

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